f(x)=2sin²(π/4+x)-√3 cos2x
=1-cos(π/2+2x) -√3 cos2x
=sin2x-√3 cos2x+1
=2sin(2x-π/3) +1
X∈[π/4,π/2],2x-π/3∈[π/6,2π/3],
sin(2x-π/3) ∈[1/2,1].
2sin(2x-π/3) +1∈[2,3].
所以x=π/4时,f(x)取到最小值2；
x=5π/12时,f(x)取到最大值3.